Dual Clifford module

$$\DeclareMathOperator\Cl{Cl}$$Let $$\Cl(V)$$ be the Clifford algebra of a quadratic vector space $$(V,Q)$$ over $$k$$, and let $$M$$ be a (left) $$\Cl(V)$$-module. The dual Clifford module is typically defined as the linear dual $$M^\ast=\operatorname{Hom}_k(M,k)$$ with the Clifford action given by defining $$g\cdot \phi$$ as $$g\cdot \phi\colon m\mapsto \phi(g^{\perp}m)$$, where $$\perp$$ is the standard antiinvolution of $$\Cl(V)$$ given by $$(v_1v_2\dotsm v_n)^\perp=v_n\dotsm v_2v_1$$. This is similar (not to say it is the same) to the dual representation of a finite group $$G$$, where the standard antiinvolution of $$kG$$ is $$g\mapsto g^{-1}$$.

Yet there is another, a priori distinct and actually more natural, definition of a dual Clifford module: one can consider $${M^\vee}=\operatorname{Hom}_{\Cl(V)}(M,\Cl(V))$$ with its natural right $$\Cl(V)$$-module structure.

This is equivalently a left $$\Cl(V)^\text{opp}$$-module structure, and since $$\perp$$ is an algebra isomorphism from $$\Cl(V)$$ to $$\Cl(V)^\text{opp}$$, we see that $$M^\vee=\operatorname{Hom}_{\Cl(V)}(M,\Cl(V))$$ carries a left $$\Cl(V)$$-module structure via $$\perp$$.

Now, one may wonder whether $$M^\ast$$ and $$M^\vee$$ are isomorphic as left $$\Cl(V)$$-modules. Equivalently, and more naturally, one is wondering whether $$M^\ast$$ and $$M^\vee$$ are isomorphic as right $$\Cl(V)$$-modules, where the right $$\Cl(V)$$-module structure on $$M^\ast$$ is given by $$\phi\cdot g\colon m\mapsto \phi(gm)$$.

For the representations of a finite group $$G$$ this is precisely what happens: $$M^\ast$$ and $$M^\vee=\operatorname{Hom}_{kG}(M,kG)$$ are isomorphic as right $$kG$$-modules. An explicit isomorphism $$M^\ast\to M^\vee$$ is given by

$$\phi\mapsto\left(m \mapsto \sum_{g\in G} \phi(g^{-1}m)g\right).$$

The inverse isomorphism is simply picking the coefficient of the unit element of $$G$$.

For Clifford algebras things should go the same way, at least over $$\mathbb{R}$$ and $$\mathbb{C}$$. Let $$(e_1,\dotsc, e_n)$$ be an orthonormal basis of $$V$$, and let $$G$$ be the subgroup of $$Pin(V)$$ consisting of the elements $$\{\pm e_1^{\epsilon_1}\dotsm e_n^{\epsilon_n}\}$$ where $$\epsilon_i\in \{0,1\}$$. Then $$\Cl(V)$$-modules are equivalently $$G$$-modules such that the element $$-1$$ of $$G$$ acts as the multiplication by $$-1$$. One can then define an isomorphism $$M^\ast\to M^\vee$$ by

$$\phi\mapsto\left(m \mapsto \frac{1}{2}\sum_{g\in G} \phi(g^{-1}m)g\right)=\left(m \mapsto \sum_{g\in G^+} \phi(g^{-1}m)g\right),$$

where $$G^+\subset G$$ is the subset $$\{e_1^{\epsilon_1}\dotsm e_n^{\epsilon_n}\}$$ of “positive” elements. Here the sums are done in $$\Cl(V)$$. To write the inverse isomorphism one notices that there exist a linear morphism $$\pi\colon \Cl(V)\to k$$ that sends $$1$$ to $$1$$ and all other elements of $$G^+$$ to $$0$$.

Now the questions (assuming all of the above is correct), are:

i) can one give the isomorphism of left Clifford modules $$M^\ast\cong M^\vee$$ in an intrinsic way, without using an orthonormal basis (this would possibly give an isomorphism over an arbitrary field $$k$$)?

ii) if the intrinsic approach fails, can one at least show that the isomorphism above is independent of the chosen orthonormal basis? (This should be a simple check, but I'm postponing it in the hope of a positive answer to i).)


Then $$\dim\Cl(V, Q) = 2^n$$; if we define the trace of an element of $$\Cl(V, Q)$$ as the trace of its left multiplication $$\tr X = \tr(Y \mapsto XY)$$ then your $$\pi : \Cl(V, Q) \to k$$ is exactly $$\pi(X) = \frac1{2^n}\tr X.$$ This can be confirmed by using an orthonormal basis. Alternatively, we could construct the canonical isomorphism $$\Cl(V, Q) \cong \Ext V$$ where then $$\pi(X) = \form{X}_0$$ where the RHS is the projection onto the scalar (i.e. grade 0) part imported from $$\Ext V$$ into $$\Cl(V,Q)$$. Briefly, we can do this by defining $$\wedge : V \times \Cl(V,Q) \to \Cl(V,Q)$$ by $$v\wedge X = \frac12(vX + \hat Xv)$$ where $$\hat X$$ is the main involution applied to $$X$$ (i.e. the involution that negates all vectors); considering the map $$V \to \Hom_k\Cl(V,Q)$$ given by $$v \mapsto v\wedge\cdot$$, the universal property of $$\Ext V$$ extends this to an algebra homomorphism $$\varphi : \Ext V \to \Hom_k\Cl(V, Q)$$, and the map $$X \mapsto \varphi(X)(1)$$ can be shown to be the desired isomorphism. This is a particular instance of Chevalley's maps $$\Cl(V, Q) \cong \Cl(V, Q+Q')$$; see the notes The Clifford algebra and the Chevalley map — a computational approach by Darij Grinberg for a general approach descending from the tensor algebra.

The bilinear form associated to $$Q$$ $$B(v, w) = \frac12(Q(v + w) - Q(v) - Q(w))$$ induces an isomorphism $$\flat : V \to V^*$$; using this to apply $$Q$$ to $$V^*$$, the universal properties of the Clifford algebras $$\Cl(V, Q), \Cl(V^*, Q)$$ extend $$\flat$$ to an algebra isomorphism $$\Cl(V, Q) \to \Cl(V^*, Q)$$. Using the natural bilinear pairing on $$\Ext V^*\times\Ext V \to k$$ $$(v^*_l\wedge v^*_{l-1}\wedge\dotsb\wedge v^*_1,\; v_1\wedge v_2\wedge\dotsb\wedge v_m) \mapsto \delta_{lm}\det\bigl(v^*_i(v_j)\bigr)_{i,j=1}^m$$ then induces a linear isomorphism $$\Ext V^* \to (\Ext V)^*$$, and composing with $$\flat$$ finally gives a linear isomorphism $$\flat' : \Cl(V, Q) \to \Cl(V, Q)^*$$; the bilinear form associated to this turns out to be exactly $$(X, Y) \mapsto \form{XY}_0.$$ We can actually get the pairing $$\Ext V^*\times\Ext V \to k$$ by considering the natural Clifford algebra on $$V^*\oplus V$$ and using the trace definition of the scalar part, but I will not digress further.

Now we define the isomorphism $$M^\vee \to M^*$$ by $$\psi \mapsto \Phi_\psi,\quad \Phi_\psi(m) = \form{\psi(m)}_0.$$ This is a homomorphism since $$\form{XY}_0 = \form{YX}_0$$ and so $$\Phi_{\psi\cdot g}(m) = \form{\psi(m)g}_0 = \form{g\psi(m)}_0 = \form{\psi(gm)}_0 = (\Phi_\psi\cdot g)(m).$$ The inverse is given by $$\phi \mapsto \Psi_\phi,\quad \Psi_\phi(m) = (g \mapsto \phi(gm))^{\sharp'}$$ where $$\sharp' = (\flat')^{-1}$$. Observe: $$\Psi_{\Phi_\psi}(m) = (g \mapsto \Phi_\psi(gm))^{\sharp'} = (g \mapsto \form{\psi(gm)}_0)^{\sharp'} = (g \mapsto \form{g\psi(m)}_0)^{\sharp'} = \psi(m).$$

Your formula in terms of $$G^+$$ comes from expressing $$\sharp'$$ using that basis; the reciprocal basis of $$G^+$$ is $$\{g^{-1} \;:\; g \in G^+\}$$.

I have no idea how to handle characteristic 2. The main issue is that there is no canonical isomorphism with $$\Ext V$$ in this case, and we can't even use the trace since $$\tr X = 0$$ for all $$X$$.

• MathJax note: a blank line outside of math mode after command definitions will produce a literal blank line in the rendered source; one must either remove the newline, or put it in math mode (so that in either case the $ that ends the preamble is on the same line as the first line of actual text). I have edited accordingly. Nov 24, 2022 at 1:40 $$\DeclareMathOperator\Cl{Cl}\DeclareMathOperator\char{char}$$Not sure about i), but at least ii) is true, so the isomorphism is canonical (at least over $$\mathbb{R}$$ or $$\mathbb{C}$$). To see this it is enough to show that the morphism $$\pi\colon \Cl(V)\to k$$ mentioned in the question is actually canonical. Indeed, if $$\char(k)=0$$ there is a canonical linear isomorphism $$\bigwedge^\bullet V\to \mathrm{Cl}(V)$$ given by $$v_1\wedge v_2\wedge\dotsb \wedge v_k\mapsto \frac{1}{k!}\sum_{\sigma\in S_k}(-1)^\sigma v_{\sigma(1)}\cdot v_{\sigma(2)}\dotsm v_{\sigma(k)}.$$ Via this isomophism, $$\Cl(V)$$ inherits a natural grading (as a vector space, not as an algebra!), whose degree zero component is the 1-dimenional subspace generated by the unit element of $$\Cl(V)$$. We have therefore a distinguished canonical isomorphism $$\Cl(V)^0\cong k$$ and the projection on the degree zero component $$\pi_0\colon \Cl(V)\to \Cl(V)^0$$ is consequently a canonical $$k$$-linear map $$\pi_0\colon \Cl(V)\to k$$. It is now a simple check to see that, for any orthogonal basis $$(e_1,e_2,\dotsc, e_n)$$ of $$V$$ the linear isomorphism $$\bigwedge^\bullet V\to \Cl(V)$$ maps $$e_{i_1}\wedge e_{i_2}\wedge \cdots e_{i_k}$$ to $$e_{i_1}\cdot e_{i_2} \cdots e_{i_k}$$, so that independently of the choice an orthonormal basis of all of the elements $$g\in G^+$$ with $$g\neq 1$$ have strictly positive degree. The a priori basis dependent map $$\pi$$ mentioned in the question is therefore nothing but the canonical projection on the the degree zero component $$\pi_0$$. (The canonical linear isomorphism $$\bigwedge^\bullet V\to \Cl(V)$$ actually exists for any characteristic different from 2, but the formula is not as nice as in the characteristic zero case: one shows that the linear isomorphism induced by the basis bijection $$e_{i_1}\wedge e_{i_2}\wedge \dotsb \wedge e_{i_k} \leftrightarrow e_{i_1}\cdot e_{i_2} \dotsm e_{i_k}$$ is independent of the choice of an orthogonal basis $$(e_1,e_2\dotsc, e_n)$$) • You divide by$k!$so it is a bit strange that you just say that the characteristic is different from$2$;-) Oct 28, 2022 at 17:23 • Argh, right! Actually it is true that there is a canonical linear isomorphism if char(k) is different from 2: one writes it in the other direction, i.e.,$Cl(V)\to \wedge^\bullet V$using an orthogonal basis and then showing it is actually basis independent; but if char(k)=0 one can more elegantly write it as$\wedge^\bullet V\to Cl(V)$with no choice at all. Thanks! Correcting now. Oct 28, 2022 at 17:56 • See my answer here which constructs the canonical isomorphism${\bigwedge}V \cong \mathrm{Cl}(V)$without resorting to orthogonal bases (though I do use them to prove the map is bijective). I also want to point out that$\pi$is actually just the normalized trace: define$\mathrm{tr}(x) = \mathrm{tr}(y \mapsto xy)$for$x \in \mathrm{Cl}(V)$. Then$\pi(x) = \tfrac1{2^n}\mathrm{tr}(x)$where$n$is the dimension of$V\$. Nov 7, 2022 at 0:29